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Vectors and Kinematics

  1. Aug 22, 2011 #1
    Well, it's the day before school and I think I'm kind of screwed since I've barely finished more than half of my AP Physics homework...and I'm really not understanding a lot of the material. Help would be greatly appreciated.

    You drive a car 1200 m to the east, then 2400 m to the north. If the trip took 3.0 minutes, what was the magnitude and direction of your average velocity in m/s?

    I do have some materials in front of me, but I was never very good at math, and I've spent hours staring at the Vectors section, and I still have a very vague idea of what's going on. I would greatly appreciate it if someone would give me a small head-start to understanding the concept behind Vectors and such. Everything just seems to complicated to me.

    I also have another question via Kinematics. There is one question on my homework that reads:
    A lunar lander is making its descent to Moon Base 1. The lander descends slowly under the retro thrust of its descent engine. The engine is cut off when the lander is 4.0 m above the surface and has a downward speed of 1.5 m/s. With the engine off, the lander is in free fall. What is the speed of the lander just before it touches the surface? The acceleration due to gravity on the moon is 1.6 m/s^2."
    I realize that there are four equations to juggle with for Kinematics, and for this one, it is
    v^2 = v0^2 + 2a(x-x0). However, I'm not too clear on where to plug in the variables, and furthermore, I'm not too sure which situations are fit for which equation. I would greatly appreciate it if someone could help me out even a little bit.
     
  2. jcsd
  3. Aug 22, 2011 #2
    For your first problem you need to find the total displacement (magnitude and direction) for the motion. Draw one vector pointing to the east and another to the south. Then draw a vector pointing from the tail of the first vector to the tip of the last vector. You need the length of this line and it's angle. What mathematical tool do you have for finding the hypotenuse of this right angle triangle?
     
  4. Aug 22, 2011 #3

    PeterO

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    In the previous post I am sure they actually meant a North vector - but here is some hints for part two.

    I wish to use the following symbols

    v = final velocity
    u = initial velocity
    a = acceleration
    s = displacement
    t = time

    We have 5 equations of motion

    v = u + at
    s = ut + ½at²
    s = vt - ½at²
    s = [(u+v)/2]*t
    v² = u² + 2as

    Each equation connects 4 of the 5 variables - a different one is "missing" from each equation.

    If we know the value of three of the quantities, we can calculate the 4th in any equation.

    I always begin a problem by writing across the top line:

    v = , u = , a = , s = , t = .

    I then fill in the gaps for the variables I know - so for the moon lander:

    v = , u = 1.5 , a = 1.6 , s = 4.0 , t = .

    next I put a question mark against the quantity I was trying to find.

    v = ? , u = 1.5 , a = 1.6 , s = 4.0 , t = .

    It is now clear that we don't know or want t, so we choose the equation without the t. You had done that correctly [with your symbols], but by writing out the known , and sought, values you should know what to do next.

    Had the question been "How long before the lander reaches the ground" the question mark would have been against t, and we would have chosen the equation without a "v" in it.


    Note: It may sound trite, but the quantity you are after is located by finding the ? mark then moving back through the text to the previous punctuation mark, then reading between those punctuation marks. It is amazing how easily people can be distracted from where the question actually is.
     
  5. Aug 22, 2011 #4
    about the guy on the moon, did he fall from rest? if so this equation works pefectly:

    (2gh)^2=v

    if he does have an initial speed, use v'^2=v^2+2ad



    about the vector problem:

    you know the pythagorean theroem, right???

    use it to sum up the two distances, (1400 and 2400) to find the hypothenuse of a right triangle. The length of this hypothenuse is the magnitude

    and the direction, well, you just use tan-1 of opposite over adjacent




    Im coursing AP physics B (i believe you are taking B and not C, right???) if you wanna talk about any content contact me
     
  6. Aug 22, 2011 #5
    Millacol: Alright, so I got a kind of sideways triangle. So basically, I use Trigonometry to find the angle and length of the angle? Sine Law, correct?

    Peter: So there are actually five equations and they all follow the basic concept of those symbols?
     
  7. Aug 22, 2011 #6

    PeterO

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    1: Since the triangle is right angles, you don't need sine rule or cosine rule.

    2: Yes. There are 5 equations - though some people never use the 3rd one I listed. [When I was at school that one was never even mentioned; I found it later.]

    writing that "v=, u=, a= , s= , t= " line as the first line of EVERY problem I solve helps to maintain focus.

    EDIT: the equations can only be applied where the acceleration is constant !!!
     
  8. Aug 22, 2011 #7
    Just use the Pythagorean theorem, then use some right angle trig to get the direction angle.
     
  9. Aug 23, 2011 #8
    Thank you very much.
    But I have one more problem that's sort of confusing too.
    It goes like "Two airplane taxi as they approach the terminal. Plane 1 taxies with a speed of 12 m/s due north. Plane 2 taxies with a speed of 9.5 m/s in a direction 20 degrees north of west. What is the magnitude and direction of the velocity of Plane 1 relative to Plane 2?
    What is the magnitude and direction of the velocity of Plane 2 relative to Plane 1?:

    So, I'm drawing to draw it out into a visual, but I'm not very successful. And I'm a little confused on what the question is trying to ask too.
     
  10. Aug 23, 2011 #9

    PeterO

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    The velocity of Plane 1 wrt plane 2 is simply the differene in their velocities. Ecah of their velocities is of cours a vector, so you subtract one vector from the other.
     
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